Cambridge International AS and A Level Mathematics Pure Mathematics 1

Integration

P1^

6

You can see that the expression is undefined at x = 0, so you need to find the

integral from a to  9    and then take the limit as a →     0    from above.

You can write: 1 2

29 2

62

9 12 9

(^1212)
(^12)
x
xx
a
a
∫a = a




=×()−()
=−
d
So as a tends to zero, the integral tends to 6, and 01
9
x
∫ dx^ =^ 6.
Notice, although the left-hand side of the curve is infinitely high, it has a finite
area.
EXERCISE 6G Evaluate the following improper integrals.
1  1
0
1
x
∫ dx^2
1
1 x^3 x
∞
∫ d^
(^3 22)
1 x
x
∞
∫ d^4
2
3
2
−∞x x
−
∫ d^
5  −
∞
∫
1
1 x^2 dx^6
6
0
4
x
∫ dx^

Finding volumes by integration

When the shaded region in figure 6.25 is rotated through 360° about the x axis,

the solid obtained, illustrated in figure 6.26, is called a solid of revolution.

In this particular case, the volume of the solid could be calculated as the difference

between the volumes of two cones (^) (using V =^13 πr^2 h), but if the line y = x in figure
6.25 was replaced by a curve, such a simple calculation would no longer be possible.
x
y
y = x
O 1 2
Figure 6.25
x
y
O
Figure 6.26